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3 years ago

Raw materials used of silicone rubber

Engineering

2 answers:

n200080 [17]3 years ago

7 0

Answer:

Silicone rubber is an elastomer (rubber-like material) composed of silicone—itself a polymer—containing silicon together with carbon, hydrogen, and oxygen. Silicone rubbers are widely used in industry, and there are multiple formulations.

Explanation:

katen-ka-za [31]3 years ago

7 0

Silicone is identified as a synthetic elastomer as it is a polymer which displays viscoelasticity – that is to say it shows both viscosity and elasticity. Colloquially people call these elastic characteristics rubber. Silicone itself is made up of carbon, hydrogen, oxygen and silicon

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technician a says an out-of-round drum can cause a pulsating brake pedal. technician b says an out-of-round drum can cause the b

denis-greek [22]

According to the question of the pulsating brake pedal, both A and B are correct.

What causes brake pulsation?

Brake pulsation is mainly caused by warped rotors/brake discs. Excessive hard braking or quick stops, which can significantly overheat the discs, are the primary causes of deformed rotors. When the discs overheat, the composition of the metal disc material changes, resulting in imperfections in the metal's surface. Hotspots are noticeable irregularities. They appear as discoloured areas of the disc material, which are often bluish or blackish in appearance. The brake pedal is the pedal which you press with your foot to slow or stop a vehicle. When the driver presses the brake pedal, the system automatically delivers the appropriate pressure required to prevent colliding with the vehicle in front.

To learn more about brake pulsation
brainly.com/question/28779956

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2 years ago

The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer

lana66690 [7]

Answer:

179000 lb

Explanation:

The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.

The specific weight of water is ρw = 62.4 lbf/ft^3

These tanks are thin walled because

D / t = 4.6 / 0.1 = 46 > 10

To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:

A = 2 * π/4 * D^2 + π * D * h

The steel volume is:

V = A * t

The specific weight is

ρ = δ * g

ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3

The weight of the steel tank is:

Ws = ρs * V

Ws = ρs * A * t

Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t

Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb

The weight of water can be approximated with the volume of the tank:

Vw = π/4 * D^2 * h

Ww = ρw * π/4 * D^2 * h

Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb

Wt = Ws + Ww = 37700 + 51800 = 89500 lb

Assuming the support holds both tanks

2 * 89500 = 179000 lb

The support must be able to carry 179000 lb

3 0

4 years ago

A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled

Vikentia [17]

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, $\mu =1.92\times 10^{- 3} Pa-s$

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

$F = \tau A$

where

$\tau$ = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = $(\tau ' +\tau '')A$

$\frac{F}{A} = \tau ' +\tau ''$

Also

F = $\frac{\mu v}{d}$

where

$\mu$ = dynamic coefficient of viscosity

Pressure, P = $\frac{F}{A}$

Therefore,

$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$

8 0

3 years ago

Using the formula XC=1/(2πfC) in your answer, how would a capacitor influence a simple DC series circuit?

Kisachek [45]

Answer:

jdjdbddifnifndjrnirnrinr

5 0

3 years ago

You are standing at the edge of the roof of the engineering building, which is H meters high. You see Professor Murthy, who is h

IRINA_888 [86]

Answer:

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Explanation:

First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:

Vf = Vi + gt

where,

Vf = Velocity of egg at the time of hitting head of the Professor

Vi = initial velocity of egg = 0 m/s (Since, egg is initially at rest)

g = acceleration due to gravity

t = time taken by egg to come down

Therefore,

Vf = 0 + gt

t = Vf/g ----- equation (1)

Now, we use 3rd euation of motion for Vf:

2gs = Vf² - Vi²

where,

s = height dropped = H - h

Therefore,

2g(H - h) = Vf²

Vf = √[2g(H - h)]

Therefore, equation (1) becomes:

t = √[2g(H - h)]/g

t = √[2(H - h)/g]

Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:

s = vt

<u>This formula describes the distance from the building that Prof Murthy must be when you release the egg</u>

5 0

3 years ago